Two envelopes problem

I was agreeing with you when I talked about convergence. You said that adding a cost would solve the problem and I agree that it does. I think you were probably thinking of a literal cost like "one cent". If you choose a cost like that then switching an infinite number of times has an infinite cost. You could instead think of the cost as a fraction of your expectation, the cost is 1% of whatever you end up getting back. Now if you switch an infinite number of times you end up having a cost of zero. That might seem counterintuitive, but recall that 1/3 is .3333 repeating. So when you sum 1/3 + 1/3 + 1/3 you get .9999 repeating. Yet 1/3 + 1/3 + 1/3 is equal to one. Infinitely close to something else is basically being the thing you are infinitely close to. Even though we never get the reward we know that the fraction is becoming infinitely close to zero. People call it "converging" when we have an infinite sequence we can sum to a real value. We know before ever seeing the value that we'll be multiplying it by zero. So we can refactor the equation to be 0*ev(switch) and then take advantage of the identity of 0x=0 to declare the result to be 0. Thus, the calculation converges to zero.

The type of issue I was referring does not occur in perfect information games.

As a practical example, consider the concept of "balancing your range" in poker. If you play poker without doing that - if you play with a purely exploitative strategy where you are only trying to maximize your EV for each hand - then your strategy will be very easily exploitable by an adaptive opponent. You will frequently end up in river situations where your opponent can deduce whether you have a strong or weak hand, so they can fold to your strong hands and bluff you out of weak hands. In contrast, if you attempt to "balance your range" - that is, consider all the subgames - then you won't end up in these situations as badly. For example, when you make a particular river bet, your opponent might deduce that you have a strong hand 70% of the time and a bluff 30% of the time (as opposed to 100% and 0%).

This issue does not exist in perfect information games. Yes, my definition of "adjusting your strategy to your opponent's strategy" was overly broad to define this issue. But if you think about the poker example, where a poker player will might make a decision like "I need to bluff with this part of my range, because I need to support my strong hands [other subgames] by having some bluffs in my range in this situation" - you won't find a corresponding example from perfect information games like chess. This class of problems is unique to imperfect-information games in which an opponent is allowed to adapt their strategy to yours. If you fix the opponent's strategy, the issue disappears. If you turn the game into a perfect-information game, the issue disappears. Both requirements must be present for this issue to exist.

I thought that Noam Brown was talking about this issue in chapter 2. He discussed a simple coin toss game where one player took a strategy, and then the other player adapted by taking the optimal (exploitative) strategy against them. Then the other player changed their strategy, and the other player again adapted their strategy. And then he described a balanced (GTO) strategy. Then he said this as a conclusion:

> This shows that a player’s optimal strategy in a subgame can depend on the strategies and outcomes in other parts of the game. Thus, one cannot solve a subgame using information about that subgame alone. This is the central challenge of imperfect-information games as opposed to perfect-information games.

I thought that this corresponds perfectly to my poker example. If it doesn't, and it means something completely different, ok, sure. I'm not a mathematician. I can't even read the notation that's used in subsequent part of the paper.

I think you understood his point very well. I just think you're making a mistake in trying to recast his claim from "imperfect information games" have this property to "this narrow subset of imperfect information games" has this property. Both imperfect games with an opponent and imperfect games without an opponent have the property of needing to play as if you are in multiple subgames, because the definition of the problem is that you don't know which subgame you are in. His claim was that the policy in one subgame could influence the EV of another subgame - and in this problem, it does. I believe you're thinking of the EV of the envelope when you think you are proving this game doesn't have that property via calculation.

To see his claim applies to this game consider the case where P(Switch)=1. Being able to calculate the EV of the envelope's contents is a bit different from solving the subgame. Here, we have two subgames E1 and E2. We can know a priori what the expected values of envelope's contents in E1 and E2 are. But if you change your policy in E1, it changes your expected value in E2. If you doubt this, remember the core of the paradox again - choose to always switch and your EV is no longer the EV of the envelopes. Ergo, the EV of the subgame E1 is dependent on the strategies and policies of another subgame, E2.

I guarantee you that your solution is going to have to incorporate a set somewhere that includes both subgames. You might forget it does, because you simplify to a scalar, but it is going to be there. In perfect information, it isn't there. In imperfect information it is.

Your opponents strategy in poker is fixed; they will always play the nash equilibrium strategy, they will never play another strategy. Their strategy is fixed. They will never change it from this setting.

You've claimed that subgame perfect play can be calculated without respect to the subgame you aren't in, because you can make a choice on the basis of the EV of the subgame you are in without respect to the subgames you aren't in.

I disagree. I think you still need to account for every subgame you are in as if you are in all of them.

Let the subgame you are in be you having KK and your opponent having AA. However, obviously - you only know that you have KK.

Therefore, you should be able to compute the strategy which is the best response to 37 suited and according to your logic it should be equal to the best response to AA. After all, you have no means of determining which subgame you are in. So you have to have the same response in both subgames.

So compute the best response for KK to AA and prove that this is also the best response to 37 suited.

However, you've claimed you don't need to calculate this with respect to other subgames. So your computation of 37 suited and your computation for AA must not be equal to each other - if they are, then you share terms. You calculated them with respect to each other.

Let Br = Best response.

Write a program which shows Br(p1, p2, I[KK]) != Br(p1, p2, I[KK]) and Br(p1, p2, I[KK]) = Br(p1, p2, I[KK]) simultaneously. (That is to say, both your policy and your opponents policy are fixed)

My contention is that you can't do this. You claim you can. You are free to use a simpler variant of poker - Kuhn poker - so that the computation becomes more tractable.